Calculus Examples

$s = - 16 + t - 12 t^{2}$ , $t = 8$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $- 16 + t - 12 t^{2}$ with respect to $t$ is $\frac{d}{d t} [- 16] + \frac{d}{d t} [t] + \frac{d}{d t} [- 12 t^{2}]$ .

$\frac{d}{d t} [- 16] + \frac{d}{d t} [t] + \frac{d}{d t} [- 12 t^{2}]$

Step 1.2

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [t] + \frac{d}{d t} [- 12 t^{2}]$

Step 1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$0 + 1 + \frac{d}{d t} [- 12 t^{2}]$

Step 2

Evaluate $\frac{d}{d t} [- 12 t^{2}]$ .

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Step 2.1

Since $- 12$ is constant with respect to $t$ , the derivative of $- 12 t^{2}$ with respect to $t$ is $- 12 \frac{d}{d t} [t^{2}]$ .

$0 + 1 - 12 \frac{d}{d t} [t^{2}]$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$0 + 1 - 12 (2 t)$

Step 2.3

Multiply $2$ by $- 12$ .

$0 + 1 - 24 t$

Step 3

Simplify.

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Step 3.1

Add $0$ and $1$ .

$1 - 24 t$

Step 3.2

Reorder terms.

$- 24 t + 1$

Step 4

Evaluate the derivative at $t = 8$ .

$- 24 \cdot 8 + 1$

Step 5

Simplify.

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Step 5.1

Multiply $- 24$ by $8$ .

$- 192 + 1$

Step 5.2

Add $- 192$ and $1$ .

$- 191$